ANSWER
[tex]r=8,a_1=0.45[/tex]
EXPLANATION
The given series is
[tex]
\sum_{k=1}^30.45(8)^{k-1}[/tex]
The k, stands for the nth term of the series.
The 3 at the top is the lasts term in the series.
We can write this in expanded form as
[tex]0.45(8)^{1-1} + 0.45(8)^{2-1} + 0.45(8)^{3-1}[/tex]
This will give us,
[tex]0.45(8)^{0} + 0.45(8)^{1} + 0.45(8)^{2}[/tex]
We can see that the subsequent terms are arrived at, by multiplying the previous terms by a constant number which is 8.
This constant number is called the common ratio, denoted by [tex]r[/tex].
Thus
[tex]r = 8[/tex]
The first term of the series is
[tex]a_1=0.45(8)^{0} [/tex]
This simplifies to
[tex]a_1=0.45[/tex]
The correct answer is C
